2
$\begingroup$

Define a random variable $X=v^2$ where $v$ is Gaussian variable with mean $m$ and variance $\sigma^2$. I am interested in whether $$\lim_{m \rightarrow \infty} (E\ln(1+X) - \ln(E(1+X))) \rightarrow0 $$

where the variance $\sigma^2$ still remains a constant.

The following is the result of difference between $E\ln(1+X)$ and $\ln(E(1+X))$ as $m$ increases exponentially.

enter image description here

We can see the difference approaches to 0 as the increase of $m$. So I guess $\ln(E(1+X))$ is the upper bound of $E(\ln (1+X))$. Is it true and how to prove or disprove it?

Thank you!

$\endgroup$
  • $\begingroup$ In what sense does your limit make sense? On the left of your limit hypothesis it looks like you've got a sequence of expectations related to a sequence of probability measures and random variables. On far right you seem to have picked a preferred random variable and probability measure, but which one is unclear because it is in terms of $m$ which is not fixed (i.e. $m\to\infty$ on the l.h.s.). $\endgroup$ – ki3i Feb 4 '15 at 13:17
  • $\begingroup$ @ki3i I have modified my question. Is it more clear for you? $\endgroup$ – NalRa Feb 4 '15 at 13:30
  • $\begingroup$ could you please say what $m$ is? $\endgroup$ – Math-fun Feb 4 '15 at 14:25
  • $\begingroup$ @Mehdi $m$ is the mean of $v$. $\endgroup$ – NalRa Feb 4 '15 at 14:50
  • $\begingroup$ Hint: Find some sufficient condition on the random variable $Z$ to ensure that $E(\log Z_m)\to0$ when $m\to\infty$, where $$Z_m=(1+m^{-1}Z)^2+m^{-2}.$$ $\endgroup$ – Did Feb 5 '15 at 14:08
1
$\begingroup$

Henry's example shows why you're seeing $\ln(E(1+X))$ as an upper bound.

Lets try to prove that this difference converges to 0

Let $Y_m:= \ln(1+X_m)$

Note that $E[1+X_m]=m^2+\sigma^2+1$ and $E[Y_m]\geq 0$ by its definition.

By Markov's Inequality:

$$P(Y_m>a)\leq \frac{E[Y_m]}{a}\implies E[Y_m]\geq aP(Y_m>a)$$

However, by Jensen's Inequality:

$$\ln(1+m^2+\sigma^2)\geq E[Y_m]$$

Combining these gives:

$$aP(Y_m>a)\leq E[Y_m]\leq\ln(1+m^2+\sigma^2)$$

Now:

$$P(Y_m>a)=P(X_m<e^a-1)=P(v_m>\sqrt{e^a-1})+P(v_m<-\sqrt{e^a-1})$$

Standardizing $v_m$ we get:

$$P(Y_m>a)=1-\Phi\left(\frac{\sqrt{e^a-1}-m}{\sigma}\right)+\Phi\left(\frac{-\sqrt{e^a-1}-m}{\sigma}\right)$$

Thus:

$$a\left[1-\Phi\left(\frac{\sqrt{e^a-1}-m}{\sigma}\right)+\Phi\left(\frac{-\sqrt{e^a-1}-m}{\sigma}\right)\right]\leq E[Y_m]\leq \ln(1+m^2+\sigma^2),\;\forall a>0$$

Lets set $a=k\ln(1+m^2+\sigma^2),k\in(0,1)$:

$$k\ln(1+m^2+\sigma^2)\left[1-\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)+\Phi\left(\frac{-(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)\right]\leq E[Y_m]\leq \ln(1+m^2+\sigma^2)$$

Taking the limit of the LHS:

$$\lim_{m\to \infty} k\ln(1+m^2+\sigma^2)\left[1-\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)+\Phi\left(\frac{-(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)\right] = k\ln(1+m^2+\sigma^2)$$

Since $\forall k \in (0,1): (m^2+\sigma^2)^{0.5k}-m=O(m)$; therefore,

$$ k\ln(1+m^2+\sigma^2)\leq E[Y_m]\leq \ln(1+m^2+\sigma^2)$$

Maximizing the lower bound by letting $k\to 1$ gives:

$$ \ln(1+m^2+\sigma^2)\leq E[Y_m]\leq \ln(1+m^2+\sigma^2) \implies \lim_{m\to \infty} E[Y_m]=\ln(1+m^2+\sigma^2)$$

$\square$

$\endgroup$
  • $\begingroup$ There is a mistake in your answer. After standardizing $v_m$, then it should follow: $$P(Y_m>a)=1-\Phi\left(\frac{\sqrt{e^a-1}-m}{\sigma}\right)+\Phi\left(-\frac{ \sqrt{e^a-1} \color{red}{+} m}{\sigma}\right)$$ $\endgroup$ – NalRa Feb 5 '15 at 8:19
  • $\begingroup$ But the reasoning process still holds by setting $a=k \ln (1+m^2+\sigma^2)$ instead of $\ln (1+m^2+\sigma^2)$, where k approaches to 1 but cannot be equal to $1$. $\endgroup$ – NalRa Feb 5 '15 at 8:24
  • $\begingroup$ @NalRa thanks for pointing that out. I fixed the signs. Yes, the logic will still work with your approach too. Either way, you get the limit by the squeeze theorem. $\endgroup$ – user76844 Feb 5 '15 at 11:46
  • $\begingroup$ Actually, I have another concern about the last step of your reasoning process, If $\lim_{m\rightarrow \infty} \frac{f(m)}{\ln(1+m^2)}=1$, can we say that $\lim f(m) = \lim (\ln (1+m^2))$? $\endgroup$ – NalRa Feb 5 '15 at 11:55
  • $\begingroup$ @NalRa See here: dl.uncw.edu/digilib/mathematics/calculus/limits/freeze/…. We could also apply the difference rule to the log of the ratio. $\endgroup$ – user76844 Feb 5 '15 at 11:59
1
$\begingroup$

This is essentially an example of Jensen's inequality as the logarithm is a concave function $f$ with the property $$f(E[X]) \ge E[f(X)]$$ resulting from the concavity of $f$.

The direction of the inequality would reverse for a convex function.

$\endgroup$
  • $\begingroup$ Quote: "I am interested in whether [some limit = 0]". $\endgroup$ – Did Feb 5 '15 at 14:03
1
$\begingroup$

Let $Y=1+X$, By Jensen's inequality, we can have $$E(Y)<\ln(E(1+X))=\ln(1+m^2+\sigma^2)$$ Since $Y=\ln (1+v^2)>0$, apply Markov inequality, for any $a>0$, $$P(Y>a)\leq \frac{EY}{a}$$ $$EY\geq aP(Y>a)$$ So we have $$aP(Y>a)\leq EY \leq \ln(1+m^2+\sigma^2)$$ Set $a=k \ln(1+m^2+\sigma^2)$, where $0<k<1$. $$P(Y>a)=1-\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)+\Phi\left(\frac{-(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)$$ So $$aP(Y>a)=k\ln(1+m^2+\sigma^2)(1-\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)+\Phi\left(\frac{-(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right))$$

Next we show that $$D=\lim_{m \rightarrow \infty} k\ln(1+m^2+\sigma^2)\left[\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)-\Phi\left(\frac{-(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)\right]=0$$

Firstly because $\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right) \geq \Phi\left(\frac{-(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)$ So $D \geq 0$.

And $D \leq \lim_{m \rightarrow \infty} \ln(1+m^2+\sigma^2)\Phi\left(\frac{(m^2+\sigma^2)^{0.5k}-m}{\sigma}\right)$

Let $M=(m^2+\sigma^2)^{0.5k}-m$, note that because $k<1$, so the order of $M$ is the same as $m$, and $M \rightarrow -\infty$ as $m \rightarrow \infty$

$$ \begin{aligned} D &\leq \frac{1}{\sqrt{2\pi}} \lim_{m \rightarrow \infty} \ln(1+m^2+\sigma^2) \int_{-\infty}^{M} e^\frac{-x^2}{2} \, dx \\ &\leq \frac{1}{\sqrt{2\pi}} \lim_{m \rightarrow \infty} \ln(1+m^2+\sigma^2) \int_{-M}^{\infty} e^{-x} \, dx \\ &=\frac{1}{\sqrt{2\pi}} \lim_{m \rightarrow \infty} \ln(1+m^2+\sigma^2)e^M=0 \end{aligned} $$

Then we have $$\lim_{m \rightarrow \infty} aP(Y>a)=k\ln(1+m^2+\sigma^2)$$ Because the choice of $k$ is arbitrary except $0<k<1$, let's choose $k$ approach to 1. Then $$\lim_{k \rightarrow 1} \lim_{m \rightarrow \infty} aP(Y>a)=\ln(1+m^2+\sigma^2)$$

$$\ln (1+m^2+\sigma^2)\leq \lim_{m \rightarrow \infty} E(Y) \leq \ln (1+m^2+\sigma^2)$$

So $$\lim_{m \rightarrow \infty} E(Y) = \ln(1+m^2+\sigma^2)$$

$\endgroup$
  • $\begingroup$ In your defintion of $D$, did you mean to have a $k$ in from of the logarithm? $\endgroup$ – user76844 Feb 5 '15 at 13:58
  • $\begingroup$ @Eupraxis1981 yes, I want to show $D=0$. $\endgroup$ – NalRa Feb 5 '15 at 14:03
  • $\begingroup$ I'm no @Did, but for what it's worth, your proof appears sound. I incorporated your insight about limiting the order of $a$ into my response as well. $\endgroup$ – user76844 Feb 5 '15 at 15:03
  • $\begingroup$ @Eupraxis1981 Actually, what I am worry about is the limit of $k$. That is when $k$ approaches to 1, whether this will put some influence on our $\lim$ on m? Can we just separate $m$ and $k$ as we assume? $\endgroup$ – NalRa Feb 5 '15 at 15:18
  • 1
    $\begingroup$ @Eupraxis1981 I understand what you mean and I come to believe that this proof is sound. Your answer is accepted and thank you for your help :) $\endgroup$ – NalRa Feb 5 '15 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.